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Copyright © 2011 Pearson, Inc.
2.4Real Zeros of Polynomial Functions
Slide 2.4 - 2 Copyright © 2011 Pearson, Inc.
What you’ll learn about
Long Division and the Division Algorithm Remainder and Factor Theorems Synthetic Division Rational Zeros Theorem Upper and Lower Bounds
… and whyThese topics help identify and locate the real zeros of polynomial functions.
Slide 2.4 - 3 Copyright © 2011 Pearson, Inc.
Division Algorithm for Polynomials
Let f (x) and d(x) be polynomials with the degree of f
greater than or equal to the degree of d, and d(x) ≠0. Then there are unique polynomials q(x) and r(x), called the quotient and remainder, such that
f (x) =d(x)⋅q(x) + r(x)where either r(x) =0 or the degree of r is less than the
degree of d.The function f (x) in the division algorithm is the
dividend, and d(x) is the divisor.If r(x) =0, we say d(x) divides evenly into f (x).
Slide 2.4 - 4 Copyright © 2011 Pearson, Inc.
Example Using Polynomial Long Division
Use long division to find the quotient and remainder
when 2x4 + x3 −3 is divided by x2 + x+1.
Slide 2.4 - 5 Copyright © 2011 Pearson, Inc.
Example Using Polynomial Long Division
x2 + x+1 2x4 + x3 + 0x2 + 0x−3
2x4 + 2x3 + 2x2
−x3 −2x2 + 0x−3
−x3 −x2 −x
−x2 + x−3
−x2 −x−1 2x−2
2x2 −x−1
2x4 + x3 −3( )÷ x2 + x+1( ) =2x2 −x−1+2x−2
x2 + x+1
Slide 2.4 - 6 Copyright © 2011 Pearson, Inc.
Remainder Theorem
If polynomial f (x) is divided by x −k,then the remainder is r = f (k).
Slide 2.4 - 7 Copyright © 2011 Pearson, Inc.
Example Using the Remainder Theorem
Find the remainder when f (x) =2x2 −x+12is divided by x+ 3.
Slide 2.4 - 8 Copyright © 2011 Pearson, Inc.
Example Using the Remainder Theorem
r = f (−3) =2 −3( )2 − −3( )+12 =33
Find the remainder when f (x) =2x2 −x+12is divided by x+ 3.
Slide 2.4 - 9 Copyright © 2011 Pearson, Inc.
Factor Theorem
A polynomial function f (x) has a factor x −kif and only if f (k) =0.
Slide 2.4 - 10 Copyright © 2011 Pearson, Inc.
Fundamental Connections for Polynomial Functions
For a polynomial function f and a real number k the following statements are equivalent:
1. x = k is a solution (or root) of the equation f(x) = 0
2. k is a zero of the function f.
3. k is an x-intercept of the graph of y = f(x).
4. x – k is a factor of f(x).
Slide 2.4 - 11 Copyright © 2011 Pearson, Inc.
Example Using Synthetic Division
Divide 3x3 −2x2 + x−5 by x−1 using synthetic division.
Slide 2.4 - 12 Copyright © 2011 Pearson, Inc.
Example Using Synthetic Division
1 3 −2 1 −5 3
1 3 −2 1 −5 3 1 2 3 1 2 −3
3x3 −2x2 + x−5x−1( )
=3x2 + x+ 2−3
x−1( )
Divide 3x3 −2x2 + x−5 by x−1 using synthetic division.
Slide 2.4 - 13 Copyright © 2011 Pearson, Inc.
Rational Zeros Theorem
Suppose f is a polynomial function of degree n ≥1of the form
f (x) =anxn +an−1x
n−1 + ...+a0 ,with every coefficient an integer and a0 ≠0. If x=p/q is a rational zero of f ,where p and q have no common integer factors
other than 1, then
g p is an integer factor of the constant coefficient a0 ,g q is an integer factor of the leading coefficient an.
Slide 2.4 - 14 Copyright © 2011 Pearson, Inc.
Upper and Lower Bound Tests for Real Zeros
Let f be a polynomial function of degree n ≥1 with apositive leading coefficient. Suppose f (x) is divided
by x−k using synthetic division.g If k≥0 and every number in the last line is nonnegative
(positive or zero), then k is an upper bound for the
real zeros of f .g If k≤0 and the numbers in the last line are alternately
nonnegative and nonpositive, then k is a lower boundfor the real zeros of f .
Slide 2.4 - 15 Copyright © 2011 Pearson, Inc.
Example Finding the Real Zeros of a Polynomial Function
Find all of the real zeros of
f (x) =2x5 −x4 −2x3 −14x2 −6x+ 36and identify them as rational or irrational.
Slide 2.4 - 16 Copyright © 2011 Pearson, Inc.
Example Finding the Real Zeros of a Polynomial Function
Potential Rational Zeros :
Factors of 36
Factors of 2:±1,±2,±3,±4,±6,±9,±12,±18,±36,
±12,±
32,±
92
Find all of the real zeros of
f (x) =2x5 −x4 −2x3 −14x2 −6x+ 36and identify them as rational or irrational.
Slide 2.4 - 17 Copyright © 2011 Pearson, Inc.
Example Finding the Real Zeros of a Polynomial Function
The graph suggests that only
x =−32, x=
32, and x=2 be
considered. Synthetic division
shows that only x=−32, and
x=2are zeros of f , and
f x( ) = 2x+ 3( ) x−2( ) x3 + 2x−6( ).
Slide 2.4 - 18 Copyright © 2011 Pearson, Inc.
Example Finding the Real Zeros of a Polynomial Function
Now let g x( ) =x3 + 2x−6.potential zeros:±1,±2,±3,±6but the graph shows that
these values are not zeros of g.So f has no more rational zeros.
Slide 2.4 - 19 Copyright © 2011 Pearson, Inc.
Example Finding the Real Zeros of a Polynomial Function
Using grapher methods with either function
f or g( ) shows that an irrational zero exists
at x≈1.456. The upper and lower bound
tests can be applied to f or g to show that
there are no zeros outside the viewing
windows shown.The real zeros of f are the rational numbers
x=−32 and x=2, and an irrational number
x≈1.456.
Slide 2.4 - 20 Copyright © 2011 Pearson, Inc.
Example Finding the Real Zeros of a Polynomial Function
Find all of the real zeros of f (x) =2x4 −7x3 −8x2 +14x+ 8.
Slide 2.4 - 21 Copyright © 2011 Pearson, Inc.
Example Finding the Real Zeros of a Polynomial Function
Potential Rational Zeros :
Factors of 8
Factors of 2:±1,±2,±4,±8
±1,±2=±1,±2,±4,±8,±
12
Compare the x-intercepts of the
graph and the list of possibilities,and decide that 4 and−1/2 arepotential rational zeros.
Find all of the real zeros of f (x) =2x4 −7x3 −8x2 +14x+ 8.
Slide 2.4 - 22 Copyright © 2011 Pearson, Inc.
Example Finding the Real Zeros of a Polynomial Function
4 2 −7 −8 14 8 8 4 −16 −8 2 1 −4 −2 0This tells us that
2x4 −7x3 −8x2 +14x+ 8 =(x−4)(2x3 + x2 −4x−2).
Find all of the real zeros of f (x) =2x4 −7x3 −8x2 +14x+ 8.
Slide 2.4 - 23 Copyright © 2011 Pearson, Inc.
Example Finding the Real Zeros of a Polynomial Function
Find all of the real zeros of f (x) =2x4 −7x3 −8x2 +14x+ 8.
−1 / 2 2 1 − 4 − 2
−1 0 2
2 0 − 4 0
This tells us that
2x4 − 7x3 − 8x2 +14x + 8 = 2(x − 4) x +1
2⎛⎝⎜
⎞⎠⎟
x2 − 2( ).
The real zeros are 4, −1
2,± 2.
Slide 2.4 - 24 Copyright © 2011 Pearson, Inc.
Quick Review
Rewrite the expression as a polynomial in standard form.
1. 2x3 + 3x2 + x
x
2. 2x5 −8x3 + x2
2x2
Factor the polynomial into linear factors.
3. x3 −16x
4. x3 + x2 −4x−4
5. 6x2 −24
Slide 2.4 - 25 Copyright © 2011 Pearson, Inc.
Quick Review Solutions
Rewrite the expression as a polynomial in standard form.
1. 2x3 + 3x2 + x
x 2x2 + 3x+1
2. 2x5 −8x3 + x2
2x2 x3 −4x+12
Factor the polynomial into linear factors.
3. x3 −16x x x+ 4( ) x−4( )
4. x3 + x2 −4x−4 x+1( ) x+ 2( ) x−2( )
5. 6x2 −24 6 x+ 2( ) x−2( )